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# How many different (eight letter) words can be formed out of the letters of the word DAUGHTER so that(a) The word starts with D and ends with R (b) Position of letter H remains unchanged (c) Relative position of vowels and consonants remain unaltered (d) No two vowels are together(e) All vowels occur together (f) All vowels never occur together Verified
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Hint: Here in the given question we are asked to find the number of words formed by the given word DAUGHTER can be arranged under a given condition. To solve this problem we use the concept of the permutation. A permutation is defined as the number of possible arrangements in a collection of items, the repetition is not allowed.

Complete step-by-step solution:
(a) Here in this sub question we are asked to find the number of words that can be formed by keeping the letters D and R placed at the starting and ending of the word. In that case the D and R place remains fixed and thus can only be arranged in 1! way each. The remaining 6 letters are A, U, G, H, T, E, R can only be arranged in 6! ways. Therefore, this can be arranged as follows
$\Rightarrow$ 1! $\times$6! $\times$ 1! =1 $\times$6 $\times$5 $\times$4 $\times$3 $\times$2 $\times$1 $\times$1 = 720
(b) From the given 8 letter word when one of the letter H positions remains unchanged the remaining 7 letters can be arranged in 7! ways, therefore 5040 words can be formed.
(c) The number of ways the word can be arranged when the relative position of vowels and consonants remain unaltered will be in one way that is DAUGHTER only.
(d) To find the number of ways a word can be arranged such that no two vowels should occur together can be found by subtracting the total number of ways the given word is arranged and when two vowels occur together.
Total number of ways the given word can be arranged is 8! ways
$\Rightarrow$8! = 8 $\times$7 $\times$6 $\times$5 $\times$4 $\times$3 $\times$2 $\times$1 = 40320------(1)
When 2 vowels are together the number of ways the word can be arranged is in 2! ways. Out of 8 letters, now 2 vowels always occur together and will be fixed and considered as one object. Considering the remaining 6 letters and 2 vowels as one object, we can arrange this in 7!2! Ways.
$\Rightarrow$ 7! $\times$2! = 7 $\times$6 $\times$5 $\times$4 $\times$3 $\times$2 $\times$1 $\times$2 = 10080------(2)
Now subtracting (1) and (2) we get
$\Rightarrow$No two vowels occur together = 8! -7! 2! = 40320 - 10080 = 30240
Therefore, we can form 30240 words in which no to vowels will be together
(e) The vowels in the given word are A, U, E. These vowels can be arranged in 3! ways. To find the number of words formed when all vowels occur together can be found by grouping the vowels as one object and remaining letters are 5, now all together these letters can be arranged in 6! ways.
$\Rightarrow$6! $\times$3! = 6 $\times$5 $\times$ 4 $\times$3$\times$2 $\times$1 $\times$3 $\times$2 $\times$1 = 4320------(3)
Thus 720 words can be formed when all vowels are together.
(f) To find the number of words when all vowels are not together, subtract the total number of words formed with the number of words formed when all the vowels are together. In the previous sub question we found the number of words formed when all vowels are together and total number of words the given word can form using this we can further proceed as follows
$\Rightarrow$ All vowels never occur together = 8! - 6! $\times$3! = 40320 – 4320 = 36000.
Therefore 36000 words can be formed by these letters with no vowel occurring together.

Note: As we see in the given question, combination means choosing elements is only that matters, whereas permutation is an ordered combination. The formula used to find combination is ${}^n{C_r} = \dfrac{{n!}}{{r! \times \left( {n - r} \right)!}}$ . Permutation is a method used to calculate the total outcome of a situation where order is important, the formula used to find permutation is and ${}^n{P_r} = \dfrac{{n!}}{{\left( {n - r} \right)!}}$.